Question

Commercially available concentrated hydrochloric acid contains 38% HCl by mass having density 1.19 g $cm^{-3}$ What volume (in cc) of concentrated hydrochloric acid is required to make 1.00 L of 0.10 M HCl?

• A. 12.38
• B. 6.19
• C. 4
• D. 8.1

Answer 1

Answer: (D)

8.1

Answer 2

1. Calculate the molarity of the concentrated HCl solution.

   • Given: 38% HCl by mass, density = 1.19 g/cm³ (or 1.19 g/mL).
   • Assume 100 g of the concentrated solution.
   • Mass of HCl = 38 g.
   • Volume of solution = $\frac{\text{Mass}}{\text{Density}} = \frac{100 \text{ g}}{1.19 \text{ g/mL}} \approx 84.03 \text{ mL}$
   • Molar mass of HCl = 1 (H) + 35.5 (Cl) = 36.5 g/mol.
   • Number of moles of HCl = $\frac{\text{Mass of HCl}}{\text{Molar mass of HCl}} = \frac{38 \text{ g}}{36.5 \text{ g/mol}} \approx 1.041 \text{ mol}$.
   • Molarity ($M_1$) = $\frac{\text{Moles of solute}}{\text{Volume of solution in Liters}}$
   • Volume of solution in Liters = $84.03 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.08403 \text{ L}$.
   • $M_1 = \frac{1.041 \text{ mol}}{0.08403 \text{ L}} \approx 12.388 \text{ M}$. We use 12.38 M for consistency with common exam values.

2. Use the dilution formula $M_1V_1 = M_2V_2$.

   • $M_1$ = Molarity of concentrated HCl = 12.38 M

   • $V_1$ = Volume of concentrated HCl required (in mL or cc) = ?

   • $M_2$ = Molarity of the desired dilute solution = 0.10 M

   • $V_2$ = Volume of the desired dilute solution = 1.00 L = 1000 mL.

   • $12.38 \text{ M} \times V_1 = 0.10 \text{ M} \times 1000 \text{ mL}$

   • $V_1 = \frac{0.10 \text{ M} \times 1000 \text{ mL}}{12.38 \text{ M}}$

   • $V_1 = \frac{100}{12.38} \text{ mL} \approx 8.0775 \text{ mL}$

   Since 1 cc = 1 mL, the volume required is approximately 8.0775 cc. The closest option is 8.1 cc.